# Anyone know electrical physics?

Discussion in 'Community Discussion' started by rye9, Sep 21, 2008.

1. ### rye9 macrumors 65816

Joined:
Sep 20, 2005
Location:
New York (not NYC)
#1
I can't seem to figure out the current across the battery in the following setup... the resistor through the center confuses me.

File size:
34.3 KB
Views:
92

Joined:
Jun 16, 2007
#2
3. ### greg555 macrumors 6502a

Joined:
Mar 24, 2005
Location:
#3
I think you are going to have to write a bunch of equations then play around with them solve for the unknowns.

Define a bunch of currents: i1 is the current through R1, i2 is the current through i2, and so on.

The current out of the battery has to equal the current returning to the battery. So i1+i2 = i4+i5.

Because of the way the 1 ohm resistors are placed (R1 and R5) some of the current through R1 will go through R3 and the rest through R4. So we write two more equations: i1 = i3+i4 and i5 = i2+i3.

You also need to write equations for the R1-4 node and the R2-5 node: V14 = 14-(1*i1) and V14 = 2*i4 and V25 = 14-(2*i2) and V25 = 1*i5.

Lastly (maybe) we know that V14 = V25+ i3*1. All that is left is a bunch of substitutions and rearranging until the answer pops out. Easy but I'm too lazy to do it.

It's easy to find the min and max that it could be. If R3 was missing (open) then there would be two 3 ohm loads on the battery. So 14/1.5 = 9.333 Amps. If R3 was shorted then R1 and R2 would parallel to 0.667 ohms and so would R4 and R4. So the load on the battery would be 1.333 ohms giving a current 10.5 ohms.

Since R3 is between 0 and infinity ohms the battery current must be between 10.5 and 9.333 Amps.

Good luck - Greg

4. ### gusious macrumors 65816

Joined:
Dec 2, 2007
Location:
Greece
#4
Old good memories from the past....!